Differential Equations

A DE relates a function $y(x)$ to its derivatives. The order of a DE is the highest derivative that appears.

First-order: involves $y'$ (but not $y''$). Example: $y' = 2xy$.

Second-order: involves $y''$. Example: $y'' + 4y = 0$ (simple harmonic motion).

A solution is a function $y(x)$ that satisfies the equation for all $x$ in some interval.

The general solution contains arbitrary constants (one constant for first-order, two for second-order). An initial value problem (IVP) provides conditions to determine these constants.

A direction field visualizes a first-order DE $y' = f(x,y)$ by drawing short line segments with slope $f(x,y)$ at many points $(x,y)$.

The solution curves must be tangent to these segments at every point. So by drawing segments, you can see the shape of solutions without solving the equation.

Direction fields build intuition: you can spot equilibrium solutions (horizontal segments), identify whether solutions grow or decay, and see the qualitative behavior before doing any algebra.

Example: for $y' = -y$, the slopes are negative when $y > 0$ and positive when $y < 0$. Solutions decay toward $y = 0$, which matches $y = Ce^{-x}$.

A first-order DE is separable if it can be written as $\frac{dy}{dx} = g(x) \cdot h(y)$, where one factor depends only on $x$ and the other only on $y$.

To solve: separate the variables: $\frac{1}{h(y)}\,dy = g(x)\,dx$. Then integrate both sides.

Don't forget the constant of integration! Usually written as $+C$ on one side.

Example: $\frac{dy}{dx} = xy$. Separate: $\frac{1}{y}\,dy = x\,dx$. Integrate: $\ln|y| = \frac{x^2}{2} + C$. Solve: $y = Ae^{x^2/2}$ where $A = \pm e^C$.

Watch for lost solutions: when dividing by $h(y)$, any value where $h(y) = 0$ might be a constant solution (equilibrium) that you lose in the algebra.

The DE $\frac{dy}{dt} = ky$ is separable with solution $y(t) = y_0 e^{kt}$ where $y_0 = y(0)$.

If $k > 0$: exponential growth (populations, compound interest, viral spread).

If $k < 0$: exponential decay (radioactive decay, cooling, drug elimination).

Half-life: the time for the quantity to halve. If $y = y_0 e^{kt}$, the half-life is $t_{1/2} = \frac{\ln 2}{|k|}$.

Doubling time: the time for the quantity to double. Doubling time $= \frac{\ln 2}{k}$ for $k > 0$.

The temperature of an object changes at a rate proportional to the difference between its temperature and the ambient temperature: $\frac{dT}{dt} = -k(T - T_{\text{env}})$.

This is separable. Solution: $T(t) = T_{\text{env}} + (T_0 - T_{\text{env}})e^{-kt}$ where $T_0$ is the initial temperature.

As $t \to \infty$, $T(t) \to T_{\text{env}}$. The object approaches room temperature exponentially.

Standard form: $\frac{dy}{dx} + P(x)y = Q(x)$. The key: this is always solvable using an integrating factor.

Integrating factor: $\mu(x) = e^{\int P(x)\,dx}$.

Multiply the entire equation by $\mu$: the left side becomes $\frac{d}{dx}[\mu(x) \cdot y]$.

Integrate both sides: $\mu(x) \cdot y = \int \mu(x) Q(x)\,dx + C$.

Solve for $y$: $y = \frac{1}{\mu(x)}\left[\int \mu(x) Q(x)\,dx + C\right]$.

Why it works: multiplying by $\mu$ turns the left side into a perfect derivative, which can be integrated directly.

The logistic equation models growth with a carrying capacity $K$: $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$.

When $P$ is small compared to $K$, growth is approximately exponential ($\approx rP$). As $P$ approaches $K$, growth slows and stops.

Solution: $P(t) = \frac{K}{1 + Ae^{-rt}}$ where $A = \frac{K - P_0}{P_0}$.

The graph is an S-shaped (sigmoidal) curve. The population grows fastest at $P = K/2$ (the inflection point).

Applications: population biology, spread of diseases, adoption of technology, logistic regression in machine learning.

A Bernoulli equation has the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ where $n \neq 0, 1$.

Trick: substitute $v = y^{1-n}$, then $\frac{dv}{dx} = (1-n)y^{-n}\frac{dy}{dx}$.

After substitution, the equation becomes linear in $v$: $\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$.

Solve the linear equation for $v$, then convert back to $y$.

Standard form: $ay'' + by' + cy = 0$ with constant coefficients.

Guess $y = e^{rx}$, substitute, and get the characteristic equation: $ar^2 + br + c = 0$.

Case 1 - Two distinct real roots $r_1, r_2$: general solution is $y = C_1 e^{r_1 x} + C_2 e^{r_2 x}$.

Case 2 - Repeated root $r$: general solution is $y = (C_1 + C_2 x)e^{rx}$. The extra $x$ factor accounts for the repeated root.

Case 3 - Complex roots $r = \alpha \pm \beta i$: general solution is $y = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x))$. This produces oscillatory behavior (springs, circuits).

Form: $ay'' + by' + cy = g(x)$ where $g(x) \neq 0$.

The general solution is $y = y_h + y_p$ where $y_h$ is the homogeneous solution (solve with $g=0$) and $y_p$ is any particular solution.

Method of undetermined coefficients: guess the form of $y_p$ based on $g(x)$.

If $g(x) = e^{kx}$, try $y_p = Ae^{kx}$. If $g(x) = \sin(kx)$ or $\cos(kx)$, try $y_p = A\cos(kx) + B\sin(kx)$. If $g(x)$ is a polynomial of degree $n$, try a polynomial of degree $n$.

If your guess duplicates a term in $y_h$, multiply by $x$ (or $x^2$ if needed).

Variation of parameters is a more general method that works for any $g(x)$ but involves more computation.

A mass-spring system obeys $my'' + cy' + ky = F(t)$ where $m$ is mass, $c$ is damping, $k$ is spring constant, and $F(t)$ is external force.

Undamped free oscillation ($c=0$, $F=0$): $y'' + \omega^2 y = 0$ with $\omega = \sqrt{k/m}$. Solution: $y = A\cos(\omega t) + B\sin(\omega t)$. Pure sinusoidal motion.

Damped oscillation ($c > 0$, $F=0$): the solution includes $e^{\alpha t}$ with $\alpha < 0$, causing the amplitude to decay over time.

Overdamped ($c^2 > 4mk$): no oscillation, just slow return to equilibrium.

Critically damped ($c^2 = 4mk$): fastest return to equilibrium without oscillating.

Underdamped ($c^2 < 4mk$): oscillation with decaying amplitude.

An IVP provides the DE plus enough initial conditions to determine all constants.

First-order: one condition, typically $y(x_0) = y_0$.

Second-order: two conditions, typically $y(x_0) = y_0$ and $y'(x_0) = v_0$.

Procedure: solve the general solution first, then plug in the initial conditions and solve for the constants.

The Existence and Uniqueness Theorem: under mild conditions, a first-order IVP has exactly one solution through any given point.